Answer:
The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
Explanation:
Given that,
Distance between the slits = 0.04 mm
Width = 0.01 mm
Distance between the slits and screen = 1 m
Wavelength = 600 nm
We need to calculate the distance between the places where the intensity is zero due to the double slit effect
For constructive fringe
First minima from center

Second minima from center

The distance between the places where the intensity is zero due to the double slit effect



Put the value into the formula



Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
Answer:
The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is
ft/ sec
Explanation:
Given:
h(t) = 25 ft/sec
x(t) = 10 ft/ sec
h(5) = 25 ft/sec . 5 = 125 ft
x(5) = 10 ft/sec . 5 = 50 ft
Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

Lets find the derivative of distance with respect to time

Substituting the values of h(t) and x(t) and simplifying we get,



=
=
ft / sec
Answer:
q₁ = + 1.25 nC
Explanation:
Theory of electrical forces
Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
Known data
q₃=5 nC
q₂=- 3 nC
d₁₃= 2 cm
d₂₃ = 4 cm
Graphic attached
The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.
For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).
The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).
The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)
Calculation of q1
F₁₃ = F₂₃

We divide by (k * q3) on both sides of the equation



q₁ = + 1.25 nC
Answer:
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